Assignment 8: Altitudes and Orthocenters
I'll start with this proof to exercise my proofing skills.
Given triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove:
(must be a typo – HE / CF should be HF / CF)
Click the GSP file to run this problems.
First, lets work on the relationship :
By the construction of the orthocenter, we know a number of characteristics of the triangles in the above figure. We have right angles at the feet of the perpendiculars and can easily calculate the area of the various triangles as ½ Base * Height. Given the ½ ratio in this area calculation, our one and 2 constants that we are trying to prove for the sum of these ratios, and the fact that we are summing three ratios together, we have a hint that the area calculations may provide insight and perhaps an expedient proof to the problem.
Let's start by summing together areas in the same format as our above conjecture. AD is the altitude of the large triangle ABC, as is BE and CF for the alternate altitudes. We know that any of these altitudes and bases will calculate the area for us, so:
Equation (a) = the Area of the Triangle ABC.
Also, we know that the three interior triangles formed with a vertex at the orthocenter and other points concurrent with the main triangle must sum to the area of the main triangle.
Equation (b) Area(BHC) + Area(AHC) + Area(AHB) = Area(ABC).
This follows a format that may give the desired equation format of …. = 1 if we divide both sides of the equation by Area(ABC), but lets first expand the area calculations for each of these smaller triangles above.
Equation (c) = the Area of the Triangle ABC.
So lets divide this equation (c ) by Area(ABC) as suggested above and at the same time utilize the corresponding form of the area calculation from equation (a) above. I'll take a shortcut for the formula and just represent this division with the somewhat improper form:
= the Area of the Triangle ABC.
= the Area of the Triangle ABC.
It's easy to see that the '2's' cancel out and the Area calculations on the right hand side reduce to 1. Also, the bases of each of the triangles cancel each other and leave the ratio of the altitudes, as desired:
which is the desired format and proof.
It is interesting that the sum of these ratios always = 1 because we know the ratio of the corresponding lengths given a centroid is always 1/3 for each of the ratios. Therefore this summed relationship holds true for a centroid as well. We may be misled into thinking that the individual ratios above (i.e HD/AD or HE/BE) would equal 1/3 but we know this cannot be the case. GSP manipulation makes it clear that the individual ratios go way out of wack, and if the individual ratios hold true for a centroid, then they can't for the orthocenter.
Now we have the relationship
Part 1 gave plenty of clues that this will be the case. These are simply the remainders of the rations from Part 1, which had a mean of 1/3, so these logically have a mean of 2/3. Again, the know relationships as proven in assignment 4 for a centroid, give good clue to this as well.
Given this, let's try a tack that builds upon this 'remainder' concept:
AH = AD – HD
BH = BE – HE
CH = CF – HF
It appears that we can introduce our new lengths into the proven relationship above in Equation (d) by simply solving the above for HD, HE, and HF and then substituting:
HD = AD – AH
HE = BE – BH
HF = CF – CH
Substituting into equation 1(d) above,
For an obtuse triangle we see the orthocenter moves outside the triangle. The obtuse triangle and the orthocenter seem to switch roles, as the ratios formed are now based from the obtuse vertex of the main triangle rather than the orthocenter. Lets test this theory out.
Using the same logic as before:
= The area of the exterior triangle
= the area of the exterior triangle
Dividing the top equation by it's corresponding pair yields:
So, we can see that these same relationships do hold true, but the Orthocenter and the triangle vertex with the obtuse angle switch roles in these relationships.
By virtue of the above demonstrations and reasoning, the sum of the rations for the corresponding altitude lengths will also be 2.
Steps done on tab 1 of the GSP file
1. Construct any triangle ABC.
2. Construct the Orthocenter H of triangle ABC.
3. Construct the Orthocenter of triangle HBC.
4. Construct the Orthocenter of triangle HAB.
5. Construct the Orthocenter of triangle HAC.
6. Construct the Circumcircles of triangles ABC, HBC, HAB, and HAC.
If we create a triangle ABC with an Orthocenter H, and then create another triangle HBC formed by the two points and Orthocenter of the original triangle, the Orthocenter of HBC (H2) will be concurrent with point A of the original triangle.
Likewise, the Orthocenter of HAB (H3) is concurrent with the point C of the original triangle, and (H4) for HAC with point B.
Note: the centers of the circumcircles are named according to the numbering of the H points for the above triangles (C1, C2, C3, C4)
What would happen if any vertex of the triangle ABC was move to where the orthocenter H is located? Where would H then be located? Click HERE for a GSP file with "move" buttons to see these animations? Explanation?
9. Construct triangle ABC, its incircle, its three excircles,
and its nine-point circle.
10. Examine the triangle formed by the points
where the extended altitudes meet the circumcircle. How is it related to the Orthic triangle? Proof? Will the relationship still hold if the original triangle is obtuse?
a. Construct your own GSP sketch.
b. Click HERE for a picture to compare with your construction.
c. Click HERE for GSP sketch to manipulate.
Construct the three altitudes AD, BE, and CF. Extend each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.
and prove your result. Click HERE for a GSP sketch to manipulate.
are extended to meet the circumcircle at points L, M, and N, respectively. Find the angles of triangle LMN in terms of the angles A, B, and C. Does your result hold only for acute triangles?
for triangle ABH.
15. Find the triangle of minimal perimeter that can be inscribed in a given triangle.
(For a start, you may want to restrict your investigation to the given triangle being acute.)